Chain rule?
- Thread starter Tom
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StintedVisions
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Perhaps try breaking it down might help?
Say you have a function sqrt{x2-5}
Break it down into f(u) and g(x)
f(u)=u1/2
g(x)=x2-5
find the derivative of both
f'(u)=1/2u-1/2
g'(x)=2x
Then, 1/2(x2-5)-1/2(2x) is your answer.
Our TA told us, take the derivative of the outside (which would be your square root), leave the inside stuff alone, and then multiply by the derivative of the inside which would be x2-5.
Hopefully this helps. I'm only in Calc I also. Maybe someone has a better way.
Say you have a function sqrt{x2-5}
Break it down into f(u) and g(x)
f(u)=u1/2
g(x)=x2-5
find the derivative of both
f'(u)=1/2u-1/2
g'(x)=2x
Then, 1/2(x2-5)-1/2(2x) is your answer.
Our TA told us, take the derivative of the outside (which would be your square root), leave the inside stuff alone, and then multiply by the derivative of the inside which would be x2-5.
Hopefully this helps. I'm only in Calc I also. Maybe someone has a better way.
The chain rule applies whenever you implicitly or explicitly simplify taking a derivative by substituting a simpler expression for the given expression.
y = f(x). For some reason, finding this derivative looks hard.
y = h(u). The derivative of this looks easy to do.
u = g(x). And the derivative of this looks easy to do. So
\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx}.\) The du in the "numerator" "cancels" the du in the "denominator." You learned about cancelling in fractions
in the third grade.
Now, formally, that "cancellation" is complete nonsense in standard modern calculus because those are not fractions, but derivatives. However, in the bad old days before rigor, they thought of them as fractions, and the chain rule came naturally when the notation was thought to refer to fractions. So as long as you do not take it literally, it is good for memorization. It is terrible for rigor. Don't use it for a proof.
y = sin(x^2)
That looks ugly
\(\displaystyle u = x^2 \implies \dfrac{du}{dx} = 2x\ and\ y = sin(u)\ and\ \dfrac{dy}{du} = cos(u) = cos(x^2) \implies\)
\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = cos(x^2) * 2x.\)
And of course you can make the chain as long as you want, provided everything "cancels."
pka
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This may not add anything to what has already been said.
Basic fact: The chain rule is all about function compositions.
Years ago as a new assistant professor, I was regularly assigned freshman honors calculus.
Being a product of a Moore method of teaching myself, I conducted a student centered course, that means the students do most of the work in the so-called lecture period. One of the oddities in that tradition is not to allow the use of the u-substitutions.
To introduce the topic of the chain rule, I would write some complicated composite function on the marker board.
For example: \(\displaystyle f(x)=\sqrt[3]{\tan(x^4)} \).
Now send a student to the board to explain exactly how to evaluate \(\displaystyle f(2)\).
What are the exact steps required to do this evaluation? This is repeated at least times.
The object was to firmly establish the idea of order of operations.
Ok now, send three different students to the board.
Apply the chain rule to those three functions: differentiate in exactly the oppsite order.
\(\displaystyle f'(x)=\frac{1}{3}{\left( {\tan ({x^4})} \right)^{\frac{{ - 2}}{3}}}\left( {{{\sec }^2}({x^4})} \right)\left( {4{x^3}} \right)\)
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Forget the "formula" entirely. Look at the function to be differentiated. Working from the outside in, working step-by-step down inside to the variable, what is the first operation that you encounter? Differentiate that operation, leaving the "insides" intact. What is the next operation? Differentiate this operation, leaving its insides intact, and multiplying the result against the previous result. Continue until you reach the variable, at which point you're done.
Example: \(\displaystyle f(x)\, =\, \sqrt{\sin(\ln(x^2))}\)
The first operation, the one "on" (or "around") everything else, is the square root. So differentiate that, leaving everything inside the square root the same:
\(\displaystyle \displaystyle{f'(x)\, =\, \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{\sin(\ln(x^2))}}\right)\, \times\, \frac{d}{dx}\, \sin(\ln(x^2))}\)
The next operation, the one "on" everything left, is the sine, so differentiate that, leaving the log expression inside the sine the same:
\(\displaystyle \displaystyle{f'(x)\, =\, \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{\sin(\ln(x^2))}}\right)\, \times\, \cos(\ln(x^2))\, \times\, \frac{d}{dx}\, \ln(x^2)}\)
The next operation is the log, so differentiate that, leaving the x^2 inside unchanged:
\(\displaystyle \displaystyle{f'(x)\, =\, \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{\sin(\ln(x^2))}}\right)\, \times\, \cos(\ln(x^2))\, \times\, \frac{1}{x^2}\, \times\, \frac{d}{dx}\, x^2}\)
The last operation, before you've drilled down to the variable, is the squaring, so finish with that:
\(\displaystyle \displaystyle{f'(x)\, =\, \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{\sin(\ln(x^2))}}\right)\, \times\, \cos(\ln(x^2))\, \times\, \frac{1}{x^2}\, \times\, 2x}\)
Now that you've reached the variable, you're done. Multiply stuff together, simplify if possible, and you're got your derivative.
Think of complicated functions as onions, and differentiate each layer as you peel it off the variable.
